Using and Manipulating Formulas
Learning Outcomes
- Recognize formulas for perimeter, area, and volume
- Recognize the formula that relates distance, rate, and time
- Rearrange formulas to isolate specific variables
- the perimeter of a rectangle of length L and width W
- [latex]P=2L+2W[/latex]
- the area of a rectangular region of length L and width W
- [latex]A=LW[/latex]
- the volume of a rectangular solid with length L, width W, and height H
- [latex]V=LWH[/latex]
- the distance [latex]d[/latex] covered when traveling at a constant rate [latex]r[/latex] for some time [latex]t[/latex]
- [latex]d=rt[/latex].
Isolate Variables in Formulas
Sometimes, we need to isolate one of the variables in a formula in order to solve for the unknown. In the carpeting example above, we found that it would take 180 square feet of carpet to cover a 12' x 15' room. We knew the dimensions of the room and solved for how many square feet of carpet we needed, But what if, instead, we had tiles that we could use to cover the floor and we knew the length of the room, but needed to know what width room could be fully tiled? Let's say you have a banquet hall with one wall that can slide to create a room of variable width. The length of the room (the length of the movable wall) is 85 feet. If you have 4,590 square feet of tile, what must the width be to ensure the entire room is fully tiled? Use the formula [latex]A=LW[/latex], but solve it for W, width. [latex]\dfrac{A}{L}=W[/latex]. Then, substitute what you know: [latex]\dfrac{4590}{85} = W = 54[/latex]. It looks like we need to slide the wall out to create a width of 54 feet in order to use all the tile. . This technique, isolating a variable of choice in any formula, is especially helpful if you have to perform the same calculation repeatedly, or you are having a computer perform the calculation repeatedly, for different values of the unknown variable.Example: isolate a variable in a formula
Isolate the variable for width [latex]w[/latex] from the formula for the perimeter of a rectangle:[latex]{P}=2\left({l}\right)+2\left({w}\right)[/latex].
Answer: First, isolate the term with w by subtracting 2l from both sides of the equation.
[latex] \displaystyle \begin{array}{l}\,\,\,\,\,\,\,\,\,\,P\,=\,\,\,\,2l+2w\\\underline{\,\,\,\,\,-2l\,\,\,\,\,-2l\,\,\,\,\,\,\,\,\,\,\,}\\P-2l=\,\,\,\,\,\,\,\,\,\,\,\,\,2w\end{array}[/latex]
Next, clear the coefficient of w by dividing both sides of the equation by [latex]2[/latex].[latex]\displaystyle \begin{array}{l}\underline{P-2l}=\underline{2w}\\\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\\ \,\,\,\Large\frac{P-2l}{2}\,\,=\,\,w\\\,\,\,\,\,\,\,\,\,\,\,w=\Large\frac{P-2l}{2}\end{array}[/latex]
You can rewrite the equation so the isolated variable is on the left side.[latex]w=\Large\frac{P-2l}{2}[/latex]
Example: isolate a variable in a formula
Use the multiplication and division properties of equality to isolate the variable b given [latex]A=\Large\frac{1}{2}\normalsize bh[/latex]Answer:
[latex]\begin{array}{l}\,\,\,\,\,\,\,\,A=\Large\frac{1}{2}\normalsize bh\\\\\left(2\right)A=\left(2\right)\Large\frac{1}{2}\normalsize bh\\\\\,\,\,\,\,\,2A=bh\\\\\,\,\,\,\,\,\,\Large\frac{2A}{h}=\frac{bh}{h}\\\\\,\,\,\,\,\,\,\,\Large\frac{2A}{h}=\frac{b\cancel{h}}{\cancel{h}}\end{array}[/latex]
Write the equation with the desired variable on the left-hand side as a matter of convention:[latex]b=\Large\frac{2A}{h}[/latex]
Use the multiplication and division properties of equality to isolate the variable h given [latex]A=\Large\frac{1}{2}\normalsize bh[/latex]Answer:
[latex]\begin{array}{l}\,\,\,\,\,\,\,\,A=\Large\frac{1}{2}\normalsize bh\\\\\left(2\right)A=\left(2\right)\Large\frac{1}{2}\normalsize bh\\\\\,\,\,\,\,\,2A=bh\\\\\,\,\,\,\,\,\,\Large\frac{2A}{b}=\frac{bh}{b}\\\\\,\,\,\,\,\,\,\,\Large\frac{2A}{b}=\frac{h\cancel{b}}{\cancel{b}}\end{array}[/latex]
Write the equation with the desired variable on the left-hand side as a matter of convention:[latex]h=\Large\frac{2A}{b}[/latex]
Think About It
Isolate the variable for height, [latex]h[/latex], from the formula for the surface area of a cylinder, [latex]s=2\pi rh+2\pi r^{2}[/latex]. In this example, the variable h is buried pretty deeply in the formula for surface area of a cylinder. Using the order of operations, it can be isolated. Before you look at the solution, use the box below to write down what you think is the best first step to take to isolate h. [practice-area rows="1"][/practice-area]Answer: Isolate the term containing the variable h by subtracting [latex]2\pi r^{2}[/latex]from both sides.
[latex]\begin{array}{r}S\,\,=2\pi rh+2\pi r^{2} \\ \underline{-2\pi r^{2}\,\,\,\,\,\,\,\,\,\,\,\,\,-2\pi r^{2}}\\S-2\pi r^{2}\,\,\,\,=\,\,\,\,2\pi rh\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]
Next, isolate the variable h by dividing both sides of the equation by [latex]2\pi r[/latex].[latex]\begin{array}{r}\Large\frac{S-2\pi r^{2}}{2\pi r}=\frac{2\pi rh}{2\pi r} \\\\\Large\frac{S-2\pi r^{2}}{2\pi r}=h\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]
You can rewrite the equation so the isolated variable is on the left side.[latex]h=\Large\frac{S-2\pi r^{2}}{2\pi r}[/latex]
Example: use a formula to convert fahrenheit to celsius
Given a temperature of [latex]12^{\circ}{C}[/latex], find the equivalent in [latex]{}^{\circ}{F}[/latex].Answer: Substitute the given temperature in[latex]{}^{\circ}{C}[/latex] into the conversion formula:
[latex]12=\left(F-32\right)\cdot\Large\frac{5}{9}[/latex]
Isolate the variable F to obtain the equivalent temperature.[latex]\begin{array}{r}12=\left(F-32\right)\cdot\Large\frac{5}{9}\\\\\left(\Large\frac{9}{5}\normalsize\right)12=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\\\left(\Large\frac{108}{5}\normalsize\right)12=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\\21.6=F-32\,\,\,\,\,\,\,\,\,\,\,\,\,\\\underline{+32\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+32}\,\,\,\,\,\,\,\,\,\,\,\,\\\\53.6={}^{\circ}{F}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]
Example: isolate a variable to convert celsius to fahrenheit
Solve the formula shown below for converting from the Fahrenheit scale to the Celsius scale for F. Then use the new formula to convert a temperature of [latex]21^{\circ}{C}[/latex] to Fahrenheit. [latex-display]C=\left(F--32\right)\cdot\Large\frac{5}{9}[/latex-display]Answer: To isolate the variable F, it would be best to clear the fraction involving F first. Multiply both sides of the equation by [latex] \displaystyle \frac{9}{5}[/latex].
[latex]\begin{array}{l}\\\,\,\,\,\left(\Large\frac{9}{5}\normalsize\right)C=\left(F-32\right)\left(\Large\frac{5}{9}\normalsize\right)\left(\Large\frac{9}{5}\normalsize\right)\\\\\,\,\,\,\,\,\,\,\,\,\,\,\Large\frac{9}{5}\normalsize C=F-32\end{array}[/latex]
Add 32 to both sides.[latex]\begin{array}{l}\Large\frac{9}{5}\normalsize\,C+32=F-32+32\\\\\Large\frac{9}{5}\normalsize\,C+32=F\\F=\Large\frac{9}{5}\normalsize C+32\end{array}[/latex]
And we can convert [latex]21^{\circ}{C}[/latex] by substituting it into the new formula: [latex-display]F=\dfrac{9}{5}(21)+32[/latex-display] [latex-display]21^{\circ}{C}=69.8^{\circ}{F}[/latex-display]Licenses & Attributions
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- College Algebra. Provided by: OpenStax Authored by: Abramson, Jay et al.. Located at: https://cnx.org/contents/[email protected]:1/Preface. License: CC BY: Attribution.